I am working on a problem with functional PCA (AKA Kosambi–Karhunen–Loève theorem), and I'm having trouble reproducing a result from a paper.
Let $ y(t) \;\; t\in [0,T]$ with mean $\mu_{fpca}(t)$ and covariance
$$k_{pca}(s,t) \equiv \text{Cov}\{Y(s), Y(t)\} = \sum_{k=1}^{\infty} \tau_k \Psi_k(s) \Psi_k(t)$$
where $\tau_1 \geq \tau_2 \geq \dots$ are eigenvalues and $\Psi_1(t), \Psi_2(t), \dots$ are the orthonormal vectors.
Our function $y(t)$ can be represented as:
$$ y(t) = \mu_{fpca}(t) + \sum_{k=1}^{\infty} \tau_k^{1/2} \beta_k \Psi_k(t) $$
where $\beta^i_k = \tau_k^{1/2}\int_{0}^T y_i(t)\Psi_k(t)dt$ is the kth principal component score.
The following result is where I get stuck. Given that the eigenvectors $\Psi_k$’s are orthonormal, the squared distance between any two square-integrable function $y_i(t)$ and $y_j(t)$ can be simplified to
$$\lVert y_i(t) - y_j(t)\rVert^2_2 = \sum_{k=1}^{\infty} \tau_k ( \beta^i_k - \beta^j_k)^2 dt$$
Any ideas on how to derive this result? I was thinking it might have something to do with the Cauchy product, but I'm not sure.